Question

two equal circles touch each other externally at C and AB is a common tangent to the circles.Then angle ACB.​

Answer 1

Answer:

The angle will be of 90 degree

Answer 2

Step by step explanation:

• Let A be on a circle woth centre O and B be the point on the circle with O' as centre. And AB be the tangent to both circles touching at A and B.

• Let the two circles touch at C.

• Let the tangent at C meet AB at N.
• Now NA and NT are tangents to the the circle with centre O andtherefore NA= NB. Sothetriangle NAC is isosceles and angles NAC = NCA = x say.
• By similar consideration NB and NT are tangents from N to circle with centre O'. So triangle NBC is isosceles with NB=NC and therefore, angles NBC = NCB = y say.
• Therefore in triangle ABC, angles A+B+C = x + y + (x+y) = 180
• Or 2(x+y) =180.
• x+y = 180/2 = 90.
• Therefore,  
• x+y = angle ACB =180/2 =90 degree.

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